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Math 347 Set-theoretic Proofs A.J. Hildebrand

Set-theoretic Proofs

Proofs of set-theoretic relations and equalities among sets are among the simplest types of proofs and therefore present anexcellent opportunity to familiarize yourself with the logical structure of a mathematical proof and to practice proofwritingin a particularly simple context.

Tools and prerequisites

You should be familiar with the basic set-theoretic operations and relations and know their precise definitions. The followingtable summarizes the key definitions and shows how to correctly unwind expressions like x A B, x A B, etc.

x A B x A or x B(1)

xA B x A and x B(2)

x A B x A and x B(3)

xA B x A or x B(4)

x A B x A and x B(5)

xA B x A or x B(6)

x A B x= (a, b) for some a A and b B(7)

A B Ifx A, then x B .(8)

A= B A B and B A.(9)

Proving a subset relation ST

Using the definition (8) of a subset relation yields a proof of the following structure:

Let x S.. . .[Logical deductions]. . .

Thereforex T.This proves that S T.

Proving a set equality S= T

By definition (9), equality between two sets S and T is equivalent to the subset relations (i) S T and (ii) T S bothbeing true. Thus, the proof ofS= T, breaks down into two parts, (i) the proof ofS T, and (ii) the proof ofTS, each

of which follows the above template.

Example 1: Proof ofA (AB) B (where A and B are arbitrary sets)

We apply the above template with A (AB) as the set S, and B as the set T.

8/10/2019 Set Theory Prof

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Math 347 Set-theoretic Proofs A.J. Hildebrand

Let x A (AB).

Then x A and x A B, by the definition of a set difference (see (5)).

By the definition of a set difference (in the negated form (6)), xA B isequivalent to xA or x B.

Therefore we have x A and (xA or x B).

Since x A, the first of the two alternatives in xA or x B is impossible, sothe second alternative must hold, i.e., x B.

Thus we have x A and x B.

Hence x B.

This proves that A (AB) B .

Remark: The reverse inclusion, B A (A B), does not hold in general. Therefore the setsA (A B) and Bare, in general, not equal. To prove this, we exhibit a counterexample: Let A= {1, 2}, B = {2, 3}. Then A B = {1}A (AB) = {2}, while B = {2, 3}. Thus, A (AB) is a propersubset ofB , but not equal to B .

To find a counterexample, a Venn diagram can be helpful. In the above case, a Venn diagram suggests that ifB overlapswith A, but is not entirely contained in A, the reverse inclusion does not hold, thus leading to the above construction. Keepin mind that, while Venn diagrams can be useful in visualizing set relations and pointing to possible counterexamples, aVenn diagram does not constitute a proof. Once you have found a likely candidate for a counterexample (such asthe sets A = {1, 2} and B = {2, 3} above), you still need to prove that this is indeed a counterexample to the relationB= A (AB) by explicitly evaluating the sets on the left and on the right and showing that they are not equal.

Example 2: Proof ofA (B C) (A B) (A C) (where A,B,Care arbitrary sets)

We apply the above template with S= A (B C) and T = (A B) (A C).

Let x A (B C).

Then x A or x B C, by the definition of a union (see (1)).

Therefore x A or (x B and x C), by the definition of an intersection (see(3)).

In the first case (i.e., the case x A), we have x A B and x A C,by the definition of a union. By the definition of an intersection it follows thatx (A B) (A C).

In the second case (i.e., the case x B and x C), we have x A Band x A C, by the definition of a union. As before, by the definition of anintersection it follows that x (A B) (A C).

Thus in either case we have x (A B) (A C). This proves that A (B C) (A B) (A C).

Remark: In this example, the reverse inclusion relation does hold, and it can be proved in much the same way as aboveTherefore, we have the set equality A (B C) = (A B) (A C), for any sets A,B, C. (This shows that the union andintersection of sets satisfy a distributive law.)